# Intuitive Explanation For Why Dependent Equations Contain No Added Information?

I've always been taught that because dependent equations contain no added information they can be deleted without effecting the solution set. Now this makes sense to me if an equation is a constant multiple of a different equation in the system because you can just cancel that constant from both sides and get the same equation.

I'm confused about the case when one equation can be represented by some linear combination of other equations in the systems. This would make it dependent and can be removed without affecting the solution, but I can't wrap my head around why the process of, for example, adding two equations together results in a dependent equation.

Consider the following system I pulled from Wikipedia's "Independent Equation" page.

$$4x + 3y = 7$$ $$x - 2y = -1$$ $$3x + 5y = 8$$

Graph Of The System




Any one equation from the system can be represented as some combination of the other equations and is dependent. Therefore one equation can be deleted from the system without effecting the solution set as should be clear from the graph.

Question: Is there an intuitive explanation as to why adding or subtracting equations results in a dependent equation and adds no new information?

• Creating new equations using algebraic rules does not alter the solution set, but can "add information": Adding the two equations $4x+5y=13$, $3x-5y=15$ results in $7x=28$ and tells you that $x=4$, which you didn't know before. – Christian Blatter Jul 22 '15 at 11:04

Start with a pair of equations $$\begin{cases} a_1x + a_2 y = c_1 \\ b_1 x + b_2 y = c_2 \end{cases}$$ Now we want to solve for $x$ and $y$, so we assume that both of these equations are true. From this it follows that $$a_1x + a_2 y + b_1x + b_2y = c_1 + c_2\,.$$ So if $(x,y)$ is a solution for our initial pair of equations, it is also a solution to the system of equations $$\begin{cases} a_1x + a_2 y = c_1 \\ b_1 x + b_2 y = c_2 \\ a_1x + a_2 y + b_1x + b_2y = c_1 + c_2 \end{cases}$$
Conversely, assume that $(x,y)$ is a solution for this system of equations . Then (trivially) $(x,y)$ is a solution for the first two equations, so $(x,y)$ is a solution for the initial equation pair.